python:使用具有多个自定义功能的 agg [英] python: use agg with more than one customized function

问题描述

我有一个这样的数据框.

I have a data frame like this.

mydf = pd.DataFrame({'a':[1,1,3,3],'b':[np.nan,2,3,6],'c':[1,3,3,9]})

   a    b  c
0  1  NaN  1
1  1  2.0  3
2  3  3.0  3
3  3  6.0  9

我想要这样的结果数据框.

I would like to have a resulting dataframe like this.

myResults = pd.concat([mydf.groupby('a').apply(lambda x: (x.b/x.c).max()), mydf.groupby('a').apply(lambda x: (x.b/x.c).min())], axis =1)
myResults.columns = ['max','min']

        max       min
a
1  0.666667  0.666667
3  1.000000  0.666667

基本上我希望每个组的 column b 和 column c 的最大和最小比率(按 column a 分组)

Basically i would like to have max and min of ratio of column b and column c for each group (grouped by column a)

是否可以通过agg来实现?我试过 mydf.groupby('a').agg([lambda x: (x.b/x.c).max(), lambda x: (x.b/x.c).min()]).它不起作用，并且似乎无法识别列名 b 和 c.

If it possible to achieve this by agg? I tried mydf.groupby('a').agg([lambda x: (x.b/x.c).max(), lambda x: (x.b/x.c).min()]). It will not work, and seems column name b and c will not be recognized.

我能想到的另一种方法是先将比率列添加到 mydf.即 mydf['ratio'] = mydf.b/mydf.c，然后在更新的 mydf 上使用 agg 就像 mydf.groupby('a')['ratio'],agg[max,min].

Another way i can think of is to add the ratio column first to mydf. i.e. mydf['ratio'] = mydf.b/mydf.c, and then use agg on the updated mydf like mydf.groupby('a')['ratio'],agg[max,min].

有没有更好的方法通过 agg 或其他函数来实现这一点?总之，我想将自定义函数应用于分组的DataFrame，并且自定义函数需要从原始DataFrame中读取多列.

Is there a better way to achieve this through agg or other function? In summary, I would like to apply customized function to grouped DataFrame, and the customized function needs to read multiple columns from original DataFrame.

推荐答案

您可以使用自定义函数来实现这一点.

You can use a customized function to acheive this.

您可以使用以下函数使用任何输入列创建任意数量的新列.

You can create any number of new columns using any input columns using the below function.

def f(x):
    t = {}
    t['max'] = (x['b']/x['c']).max()
    t['min'] = (x['b']/x['c']).min()
    return pd.Series(t)

mydf.groupby('a').apply(f)

输出:

        max       min
a                    
1  0.666667  0.666667
3  1.000000  0.666667

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